Muddiest Point.

I thought the muddiest part of the class on Thursday was the second part of the proof for showing that sup(ca) = csup(A) . The reason why I thought this was the muddiest point is because it is easy to confuse the difference between the upper-bound and the least upper-bound of a set and that once you have shown that something is the upper-bound of a set it is easy to forget that you must show that it is the least upper-bound.
The reason why there is a second part of this proof is that in the first part we have only shown that cs is an upper-bound of set cA, where s = sup(A) and c \geq 0. In order to complete the proof we must show that cs is the least upper bound which as the name indicates means that cs is less than all other upper-bounds for the set cA. This is the difference between an upper-bound and a least upper-bound. Once we know that we must show that cs is a least upper-bound we start by letting d be and upper-bound for cA. Now we have two different cases. One where c = 0 and the other where c> 0 because as you will see the mathematics are different for each case. For the case that c>0 we know that d\geq a \forall a \in A. Then by dividing by c we can see that \frac{d}{c} \geq a \forall a \in A. This means that \frac{d}{c} is an upper-bound for A which means s \neq \frac{d}{c} since we already know s is the least upper-bound of A. Now with the inequality s \neq \frac{d}{c} we can multiply both sides by c and we see that cs \neq d. Now we know that cs is the least upper-bound if c>0, so now we must do the case that c = 0. If c = 0 then cs = 0. From here it is easy to deduce that d \geq cs because d \geq ca = 0 = cs.Now it is apparent why two cases are needed. As you can see the case where c>0 cannot be done if c could be zero as that would involve dividing by zero. In addition to this the case where c = 0 is much simpler due to the nature of multiplying by zero.

 

Posted on by Abraham Schroeder
Categories: Muddiest Point

One Response

  1. Jeremy LeCrone says:
    September 5, 2017 at 12:56 pm

    Good explanation of this aspect of the lecture. I am very happy that you went beyond simply identifying why it was confusing and gave a synopsis of the proof itself. Thank you Abe!

    There are three typos in your proof I want to point out, however, just to avoid confusion.
    1) “For the case c > 0 we know that d \geq a \ \forall a \in A…” should read “… we know that d \geq ca \ \forall c a \in cA…” (since d is an upper bound of cA, not A).
    2 and 3) “… which means that s \neq \frac{d}{c}…” and “… with the inequality s \neq \frac{d}{c}…” should both contain the inequality s \leq \frac{d}{c}.

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